Radius of Gyration Class 11
As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we define a new parameter known as the radius of gyration

$J=I \omega$
I- Moment of inertia with respect to axis of rotation
$\omega$ - Angular velocity due to angular momentum
J- The moment of momentum which is causing rotational motion.
The rate of change of angular momentum is equal to the torque applied on the body.
$$
\vec{\tau}=\frac{d \vec{J}}{d t}
$$
In rotational motion angular momentum has equal importance as linear momentum in linear motion
If torque acting of a particle is zero then
$\vec{\tau}=0 \Rightarrow \frac{d \vec{J}}{d t}=0 \quad$ or $\quad \vec{J}=$ constant
Which implies that the angular momentum remians conserved when no external torque acts on the body.

**.**It is the distance whose square when multiplied with the mass of the body gives the moment of inertia of the body about the given axis. For a body of mass $M,$ if $K$ is radius of gyration $\mathrm{I}=\mathrm{MK}^{2}$ $\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}} \quad \mathrm{K}=\sqrt{\frac{\sum_{i=1}^{n} m_{i} r_{i}^{2}}{M}}$ Through this concept a real body (particular) is replaced by a point mass for dealing its rotational motion. e.g., in case of a solid sphere rotating about an axis through its centre of mass. $\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{(2 / 5) \mathrm{MR}^{2}}{\mathrm{M}}}=\mathrm{R} \sqrt{\frac{2}{5}}$ So instead of solid sphere we can assume a point mass M at a distance $(\mathrm{R} \sqrt{\frac{2}{5}})$ from the axis of rotation for dealing the rotational motion of the solid sphere.**K depends on**(i) axis of rotation(ii) distribution of mass of body**K does not depend on**(i) Mass of the body $\quad$ (ii) Position of body (iii) On all angular physical quantity**radius of gyration does not depends on the mass of rigid body.**Graphical relation between log K and log I.

**Angular Momentum**The moment of linear momentum of a moving particle with respect to axis of rotation is known as angular momentum. It is a vector quantity, which is often represented by $\overrightarrow{\mathrm{L}}$ or $\overrightarrow{\mathrm{J}}$. It is an axial vector (i.e. always perpendicular to the plane of motion) $\begin{aligned} \text { Angular momentum } & \vec{J}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}} \\ &=\overrightarrow{\mathrm{r}} \times(\mathrm{m} \overrightarrow{\mathrm{v}})=\mathrm{m}(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{v}}) \end{aligned}$ or $\quad \vec{J}=r p \sin \theta \hat{n}$ $\theta$ is angle between $\vec{r}$ and $\vec{v}$ $\hat{n}$ is unit vector perpendicular to plane of $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{v}}$ The direction of angular momentum is perpendicular to the plane of $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{v}}$ and it given by right hand screw rule. $\begin{array}{ll}{\text { If } \theta=0^{\circ},} & {J=0 \quad \text { (Minimum) }} \\ {\text { If } \theta=90^{\circ},} & {J=m \vee r \quad \text { (Maximum) }}\end{array}$ S.I. Unit of $J$ is Jule $\times$ sec (same as that of Planck's const.) Dimension: $\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-1}$ If direction of rotation is anticlockwise, angular momentum is taken positive and if direction of rotation is clockwise, angular momentum is taken negative. The angular momentum of a system of particles is equal to the vector sum of angular momentum of each particle $\vec{J}=\vec{J}_{1}+\vec{J}_{2}+\vec{J}_{3}+\ldots \ldots \ldots . \quad \vec{J}=\sum_{i=1}^{n} \vec{J}_{i}$**Relation between angular momentum and angular velocity**

$J=I \omega$
I- Moment of inertia with respect to axis of rotation
$\omega$ - Angular velocity due to angular momentum
J- The moment of momentum which is causing rotational motion.
The rate of change of angular momentum is equal to the torque applied on the body.
$$
\vec{\tau}=\frac{d \vec{J}}{d t}
$$
In rotational motion angular momentum has equal importance as linear momentum in linear motion
If torque acting of a particle is zero then
$\vec{\tau}=0 \Rightarrow \frac{d \vec{J}}{d t}=0 \quad$ or $\quad \vec{J}=$ constant
Which implies that the angular momentum remians conserved when no external torque acts on the body.
**LAW OF CONSERVATION OF ANGULAR MOMENTUM**If no external torque is acting upon a body rotating about an axis, then the angular momentum of the body remains constant that is $\vec{J}=\overrightarrow{\mathrm{I} \omega}=\mathrm{constant}$ Proof: For a rigid body, rotating about a given axis. $\quad \vec{\tau}=1 \vec{\alpha}=\mathrm{I} \cdot \frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}$ Now, for a rigid body, rotating about a given axis, Iremains constant. $\vec{\tau}=\mathrm{I} \cdot \frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}=\frac{\mathrm{d}(\overrightarrow{\mathrm{I} \omega})}{\mathrm{dt}}=\frac{\mathrm{d} \overrightarrow{\mathrm{J}}}{\mathrm{dt}}$ If the external torque acting on the body, $\vec{\tau}=0$ $$ \frac{d \overrightarrow{\mathrm{J}}}{d t}=0, \text { and hence } \overrightarrow{\mathrm{J}}=\text { constant. } $$ For the equation $\mathrm{J}=\mathrm{I} \omega,$ it can be seen that if $\mathrm{J}$ remains constant, $\omega$ increases when I decreases and vice versa. If I changed to $\mathrm{I}_{1}$, then, in the absence of an external torque, $\omega$ will change to $\omega_{1}$, such that $\mathrm{I}_{\mathrm{i}} \omega_{\mathrm{i}}=\mathrm{I}_{\mathrm{f}} \omega_{\mathrm{f}}$ This is the law of conservation of angular momentum. Introduction to Rotational Dynamics Moment of Inertia Moment of Inertia: Perpendicular and Parallel axis theorem Radius of Gyration Law of Conservation of Angular Momentum Conservation of Angular Momentum Examples Kinetic Energy of a Rotating Body Work done in rotatory Motion Rotational Power Combine Translational and Rotational Motion Rolling without slipping Rolling on a plane surface Rolling on a Inclined Plane For Latest updates, Subscribe our Youtube ChannelClick here to get exam-ready with eSaral

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